Computing with Nonstandard Domain and Codomain Bases

Representing a Map in Two Different Bases

A matrix representation depends not only on a linear map but also on the bases used to describe its inputs and outputs. If the domain and codomain use different ordered bases, the columns still record images of domain basis vectors, but each image must be expressed in the codomain basis. Using standard coordinates for one side by mistake can give a matrix with the right shape and the wrong meaning.

Throughout, the scalar field is \(\mathbb{R}\), and all spaces considered here are finite-dimensional. Let \(V\) have ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\), and let \(W\) have ordered basis \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\). For \(\mathbf{v}\in V\), its coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in its unique expansion in \(\mathcal{B}\). The same convention defines \([\mathbf{w}]_{\mathcal{C}}\) for \(\mathbf{w}\in W\).

The arrow in the subscript records the direction: coordinates in \(\mathcal{B}\) go in, and coordinates in \(\mathcal{C}\) come out. To construct the matrix, apply \(T\) to each domain basis vector, then write each result as a combination of the codomain basis vectors. The coefficients of the \(j\)th image, in order, form column \(j\).

Proof. Write \([\mathbf{v}]_{\mathcal{B}}=(a_1,\dots,a_n)^{\top}\). By the definition of coordinates, \(\mathbf{v}=\sum_{j=1}^n a_j\mathbf{b}_j\). Linearity gives \(T(\mathbf{v})=\sum_{j=1}^n a_jT(\mathbf{b}_j)\). Taking \(\mathcal{C}\)-coordinates preserves sums and scalar multiples: if \(T(\mathbf{b}_j)=\sum_{i=1}^m d_{ij}\mathbf{c}_i\), then \(T(\mathbf{v})=\sum_{i=1}^m(\sum_{j=1}^n d_{ij}a_j)\mathbf{c}_i\). Thus the \(i\)th output coordinate is \(\sum_{j=1}^n d_{ij}a_j\), exactly the \(i\)th entry of the product of the matrix with columns \([T(\mathbf{b}_j)]_{\mathcal{C}}\) and the column \((a_1,\dots,a_n)^{\top}\). This proves both the column construction and the stated coordinate rule. \(\square\)

Compute the Image Columns, Then Apply the Matrix

The matrix acts on a coordinate vector, not directly on an unlabelled vector. In this example, the column \((4,1)^{\top}\) means \(4\mathbf{b}_1+\mathbf{b}_2\); it is not the standard-coordinate column of the input. Likewise, the output \((11,2)^{\top}\) describes coefficients of \(\mathbf{c}_1,\mathbf{c}_2\), not standard coordinates.

A Rectangular Example with Different Bases

Converting Through Standard Coordinates

When \(V=\mathbb{R}^n\) and \(W=\mathbb{R}^m\), there is another route to the same matrix. Let \(\mathbf{A}\) be the standard matrix of \(T\), so \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) for standard-coordinate columns \(\mathbf{x}\). Define the basis matrices \(\mathbf{P}_{\mathcal{B}}=\begin{bmatrix}\mathbf{b}_1&\cdots&\mathbf{b}_n\end{bmatrix}\) and \(\mathbf{P}_{\mathcal{C}}=\begin{bmatrix}\mathbf{c}_1&\cdots&\mathbf{c}_m\end{bmatrix}\). Each basis matrix converts coordinates in its named basis into standard coordinates: \(\mathbf{x}=\mathbf{P}_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\), and \(\mathbf{y}=\mathbf{P}_{\mathcal{C}}[\mathbf{y}]_{\mathcal{C}}\).

Proof. A basis matrix has the basis vectors as its columns. Since those columns are a basis, every standard-coordinate vector has a unique coefficient column in that basis. Thus \(\mathbf{P}_{\mathcal{B}}\) converts \(\mathcal{B}\)-coordinates to standard coordinates, and its inverse converts standard coordinates back to \(\mathcal{B}\)-coordinates; the same reasoning applies to \(\mathbf{P}_{\mathcal{C}}\). For any \(\mathbf{v}\in V\), its standard-coordinate column is \(\mathbf{P}_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Applying the standard matrix gives standard output coordinates \(\mathbf{A}\mathbf{P}_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Converting that output to \(\mathcal{C}\)-coordinates gives \(\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). By the coordinate rule in the preceding theorem, the matrix multiplying \([\mathbf{v}]_{\mathcal{B}}\) is precisely \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\). \(\square\)

Common Pitfall: Mixing Coordinate Systems

A frequent error is to compute \(T(\mathbf{b}_j)\) in standard coordinates and place those entries directly into the represented matrix, even though the requested codomain basis is nonstandard. In the first example, the standard-coordinate images are \((3,0)\) and \((1,2)\). The incorrect matrix would therefore be \(\begin{bmatrix}3&1\\0&2\end{bmatrix}\). Applying it to the input \((4,1)^{\top}\) gives \((13,2)^{\top}\), but this column is being interpreted as \(\mathcal{C}\)-coordinates; it would describe \(13(1,0)+2(1,1)=(15,2)\), not the actual output \((13,2)\).

The correction is to convert each image to codomain coordinates before inserting it as a column. In the example, \((3,0)\) has \(\mathcal{C}\)-coordinates \((3,0)^{\top}\), while \((1,2)\) has \(\mathcal{C}\)-coordinates \((-1,2)^{\top}\). The correct matrix is \(\begin{bmatrix}3&-1\\0&2\end{bmatrix}\), and it produces output coordinates \((11,2)^{\top}\), which describe \((13,2)\) in \(\mathcal{C}\).

Before using a computed matrix, check both labels: its input must be a coordinate column in the domain basis, and its output must be a coordinate column in the codomain basis. If a standard matrix is available, the formula \(\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}\) provides an independent check. The order matters: first convert the input to standard coordinates, apply the map, then convert the result to codomain coordinates.